How do you simplify #(216/729)^(2/3)#?

Question

3785 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-01T22:08:15

#4/9#

It is a real advantage to know all the powers up to 1000 by heart.

We would then know at once that #" "216 = 6^3 and 729 = 9^3#

Else you have to find the prime factors first.

An index that is a fraction shows both a root and a power.

#x^(p/q) = rootq(x^p) = (rootq x)^p#

#(216/729)^(2/3) = (root3 (216/729))^2#

=#(6/9)^2 " = "(2/3)^2#

=#4/9#